5.3 The Fundamental Theorem of Calculus/53: Difference between revisions

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\begin{align}
\begin{align}


\frac{d}{dx}[g(x)] = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1} \\[2ex]
\frac{d}{dx}[g(x)] &= \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right] \\[2ex]


=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1} \\[2ex]
&=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1} \\[2ex]


=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1} \\[2ex]
&=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1} \\[2ex]
 
&=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1} \\[2ex]


\end{align}
\end{align}
</math>
</math>

Revision as of 22:24, 6 September 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g(x)=\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du}


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}{\frac {d}{dx}}[g(x)]&={\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]\\[2ex]&=3\cdot {\frac {(3x)^{2}-1}{(3x)^{2}+1}}-2\cdot {\frac {(2x)^{2}-1}{(2x)^{2}+1}}\\[2ex]&=3\cdot {\frac {9x^{2}-1}{9x^{2}+1}}-2\cdot {\frac {4x^{2}-1}{4x^{2}+1}}\\[2ex]&={\frac {3(9x^{2}-1)}{9x^{2}+1}}-{\frac {2(4x^{2}-1)}{4x^{2}+1}}\\[2ex]\end{aligned}}}